Series test (Calculus 2)

Question

∞

∑ (ln^3)/n

n=1

Does it converge or diverge by what series test? Please show how you solve it please.

Answer 1

What tests have you learned so far? I presume you've learned the two comparison tests:

Which way would an inequality point if it is between the terms given (ln n)^3/n and the terms of 1/n? For this question, assume n >= 3.

What do you know about the series made from the terms of 1/n? That is, does the series ∑1/n converge or diverge?

Answer 2

Hi Mustafa! Because the series notation that you wrote is a little ambiguous (did you mean (ln(n))^3 or ln(n^3) or something???), here, I will make an assumption that you meant to say (ln(n))^3. Please let me know if you meant to say a different series. Without further ado, here I go!

We can use a limit comparison test here. Note that how the series in the question has the natural log part divided by n. This can sound similar to the series 1/n, which we know it DIVERGES. Let the nth term in the series in the question be called a sub n, and let the nth term in the series of1/n be called b sub n. Hence, we make one of the two conclusions:

1) If lim (n → ∞) (a sub n) / (b sub n) = ∞, then because b sub n DIVERGE, then we know that a sub n MUST DIVERGE.

2) If lim (n → ∞) (a sub n) / (b sub n) = D (let 0 < D < ∞, in other words, D is a positive noninfinite number), then since b sub n DIVERGES, a sub n ALSO MUST DIVERGE.

Note that how regardless of what number the result will be, the aeries in the question will DIVERGE. Hence, the series DIVERGES! <<<<(ANSWER).Let me know if you need more assistance!